My project involves robots, and the problem i'm faced with is power distributed in the robot from a 3.7 volt and 750 mAh battery. With a power booster basic to convert the 3.7 volts to 5 volt for the arduino to properly work.
The robots function well when connected to the computer via Usb, but using a fully charged battery the robot is sluggish, even the color sensor's leds flicker.
The components on the robot are the Arduino nano, One powerbooster 500 basic from adafruit, a 3.7 volt and 750 mah battery, two fitech micro servos (continue rotation), one TCS230 color sensor, and an array of 8 infrared emitters and detectors on another board on top. The infrared emitters and detectors are powered through a npn transistor, and each emitter has a 330 ohm resistor, while detectors have 10k resistor. The battery is only connected to the arduino, while the arduino sends power to the other components.
I feel the problem deals with the milliamps per hour, and how it is not enough to power every component sufficiently. I could use a npn transistor to amplify the current, even the voltage if that would help. Maybe a transistor would be too small and weak, maybe a mosfet would be sufficient (with heatsink). I'm looking to more options to solve this problem, which is why I look to this forum. Leave any opinions and comments on possible solutions.
(This may sound extremely newbie of me, but I'm pretty much a novice in electronics as of now.)
You do not have a 3.7 mAh battery. If you do, it must be tiny. Batteries normally have capacities of hundreds or thousands of mAh.
You probably have a 3.7v battery (that's the voltage per cell for most rechargeable lithium batteries - 3.7 to 4.2, depending on if you're talking nominal voltage or what you charge it up to)
Recheck specs on battery.
You cannot use a transistor to increase the capacity of a battery. It is useful for switching loads too large for the arduino to switch, but if the battery can't provide power to switch, all the transistors in the world won't help you - per first law of thermodynamics (conservation of energy)
Yes, it is a 3.7 volt and 750 mah battery. My bad for this typing error :).
DrAzzy:
You cannot use a transistor to increase the capacity of a battery. It is useful for switching loads too large for the arduino to switch, but if the battery can't provide power to switch, all the transistors in the world won't help you - per first law of thermodynamics (conservation of energy)
Well, the more you know :), Thanks for telling me.
So if a transistor won't work, and I checked the Power Booster Basic, I bought it for a reason, so that it converts dc/dc and steps it up, then what will? I checked the specs. It turns the little 3.7 volt lipoly battery to 5 volt output and more than 900 mah of power. So what gives? I'm wondering that the current is too low for the robot to consume and properly operate. Does the components from the robots altogether consume 900 mah? I can't seem to figure this out .
You can draw, in theory, a lot of current from a battery.
5000mA (5Amp) may be possible from your 750mAh battery.
There is ofcourse a practical limit for the relatively small battery you have.
mAh has 'time' in in (the 'h' for hours). How long things last.
If your battery is 750mAh, and you constantly draw 750mA, your battery would last 1hour.
If you only draw 75mA, it would last 10hours.
(ignoring practical losses for clarity)
No free lunch...
If you use a boost converter to increase the battery voltage to 5volt, and the total system is using 500mA@5volt, the boost converter will draw more current from the battery. The step-up ratio. So for 5volt/500mA >> 5/3.7 *500 = 675mA. Plus converter losses (10-20%) = ~800mA.
Use battery power wisely:
You could use two IR emitters in series, with a ~200ohm current limiting resistor.
That will save you 40mA.
Leo..
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